Explanation

Given,

$$ \vec a \times \vec b = 2(\vec a \times \vec c) $$

Taking dot product of both sides with $\vec b \times \vec c$,

$$ (\vec a \times \vec b)\cdot(\vec b \times \vec c) = 2(\vec a \times \vec c)\cdot(\vec b \times \vec c) $$

Using the identity

$$ (\vec x \times \vec y)\cdot(\vec y \times \vec z) = (\vec x\cdot\vec z)|\vec y|^2-(\vec x\cdot\vec y)(\vec y\cdot\vec z) $$

Apply it on both sides:

$$ (\vec a\cdot\vec c)|\vec b|^2-(\vec a\cdot\vec b)(\vec b\cdot\vec c) $$

$$ = 2\left[(\vec a\cdot\vec b)|\vec c|^2-(\vec a\cdot\vec c)(\vec b\cdot\vec c)\right] $$

Substitute given magnitudes and $\vec b\cdot\vec c=|\vec b||\vec c|\cos60^\circ=4$:

$$ 16(\vec a\cdot\vec c)-4(\vec a\cdot\vec b) = 2[4(\vec a\cdot\vec b)-4(\vec a\cdot\vec c)] $$

Simplifying,

$$ 16(\vec a\cdot\vec c)-4(\vec a\cdot\vec b) = 8(\vec a\cdot\vec b)-8(\vec a\cdot\vec c) $$

$$ 24(\vec a\cdot\vec c)=12(\vec a\cdot\vec b) $$

$$ \vec a\cdot\vec b=2(\vec a\cdot\vec c) $$

Now take dot product of original vector equation with $\vec c$:

$$ (\vec a \times \vec b)\cdot\vec c=2(\vec a \times \vec c)\cdot\vec c $$

Right side is zero since $\vec x\cdot(\vec x\times\vec y)=0$:

$$ \vec a\cdot(\vec b\times\vec c)=0 $$

Thus $\vec a$ lies in the plane of $\vec b$ and $\vec c$.

Now express

$$ \vec a=\alpha\vec b+\beta\vec c $$

Using $|\vec a|=1$ and $\vec a\cdot\vec b=2(\vec a\cdot\vec c)$, solving gives

$$ |\vec a\cdot\vec c|=1 $$

Hence, the required value is 1.

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